Veritasium - Firing bullet in block - along center and away from cente

[cupid.callin]

This question is about this video on YouTube, in which a bullet is fired vertically into the center of a wooden block from below, sending the block up into the air. Next, a bullet is fired vertically but off-center into a similar block from below, again causing the block to rise into the air, but simultaneously to rotate. The video asks for a prediction as to which block will rise higher.

Please also post your initial guess along with answer.

My first guess was that air friction decreases when object is rotating, but now I think that's not true (is it?).

Someone please shed some light on this.

Is it possible that the first bullet went deeper inside the first block and thus the first block had lesser kinetic energy than expected?

 

[tiny-tim]

hi cupid.callin!

hint: consider momentum, consider energy

(i think you can ignore air resistance)

 

[cupid.callin]

I know that momentum will be conserved and this in both cases linear momentum will be same after and before the collision, But I am having second thoughts about this idea, is linear momentum conserved even if angular momentum is present?

I mean, suppose that energy L is lost due to inelastic collision. So for first case,

[itex]PE_i + KE_i = PE_f + KE_f + L [/itex]

Suppose that just before collision, Center of mass is at zero potential level

[itex]K = P + L [/itex]

For second case,

[itex]PE_i + KE_i = PE_f + KE_{f,translational} + KE_{f,rotational} + L[/itex]

and if we consider same as above,

[itex]K = P + KE_f,rotational + L[/itex]

______________________

Then only explanation I can think of is that L is not same in both cases and that bullet goes deeper in first case

 

[tiny-tim]

hi cupid.callin!

I know that momentum will be conserved and this in both cases linear momentum will be same after and before the collision, But I am having second thoughts about this idea, is linear momentum conserved even if angular momentum is present?

yes, linear momentum is always conserved in collisions

(in any direction in which there is no external impulse)

Then only explanation I can think of is that L is not same in both cases and that bullet goes deeper in first case

i think you can assume that the energy loss is the same in both cases

ok, you know that the initial momentum (after the collision) is the same in both cases

so you know that the initial energy (after the collision) is also the same in both cases

sooo … ​

 

[haruspex]

i think you can assume that the energy loss is the same in both cases

No. There is less loss of energy in the off-centre case. That is how it is possible for the linear speed to be the same yet rotation also to occur.

 

[Dick]

No. There is less loss of energy in the off-centre case. That is how it is possible for the linear speed to be the same yet rotation also to occur.

Good answer! Energy loss due to friction MUST be less in the off-center collision. Momentum conservation demands it.